題目：

There are a total of n courses you have to take, labeled from 0 to n - 1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

click to show more hints.

1. This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

鏈接： http://leetcode.com/problems/course-schedule/

6/28/2017

有向圖，先求indegree，然後每次從queue當中取出來indegree為0的課程，最後看queue當中是否為空。

Kahn‘s Algorithms

 1 public class Solution {
 2     public boolean canFinish(int numCourses, int[][] prerequisites) {
 3         // calculate indegree
 4         int[] indegree = new int[numCourses];
 5         List<List<Integer>> graph = new ArrayList<>();
 6         for (int i = 0; i < numCourses; i++) {
 7             graph.add(new ArrayList<Integer>());
 8         }
 9         for (int i = 0; i < prerequisites.length; i++) {
10             indegree[prerequisites[i][0]]++;
11             graph.get(prerequisites[i][1]).add(prerequisites[i][0]);
12         }
13         
14         Queue<Integer> q = new LinkedList<Integer>();
15         // add indegree == 0 vertex to queue
16         for (int i = 0; i < numCourses; i++) {
17             if (indegree[i] == 0) {
18                 q.add(i);
19             }
20         }
21 
22         int counter = 0;
23         while (!q.isEmpty()) {
24             int vertex = q.poll();
25             counter++;
26             // decrease all adjacent vertex indegree by 1
27             int adjacentVertexLength = graph.get(vertex).size();
28             for (int i = 0; i < adjacentVertexLength; i++) {
29                 int w = graph.get(vertex).get(i);
30                 indegree[w]--;
31                 // check if new indegree is 0
32                 if (indegree[w] == 0) {
33                     q.add(w);
34                 }
35             }
36         }
37         
38         return counter == numCourses;
39     }
40 }

這道題很久之前做的，思路已經忘記了，具體題解可以參考：

http://www.cnblogs.com/yrbbest/p/4493547.html

求Course Schedule，等同問題是有向圖檢測環，vertex是course， edge是prerequisite。我覺得一般會使用Topological Sorting拓撲排序來檢測。一個有向圖假如有環則不存在Topological Order。一個DAG的Topological Order可以有大於1種。 常用的Topological Sorting算法有兩種

1. Kahn‘s Algorithms (wiki)： BFS based， start from with vertices with 0 incoming edge，insert them into list S，at the same time we remove all their outgoing edges，after that find new vertices with 0 incoming edges and go on. 詳細過程見Reference裏Brown大學的課件。
2. Tarjan‘s Algorithms (wiki)： DFS based， loop through each node of the graph in an arbitrary order，initiating a depth-first search that terminates when it hits any node that has already been visited since the beginning of the topological sort or the node has no outgoing edges (i.e. a leaf node). 詳細過程見Reference裏 NYU的課件。

別人的解法

https://discuss.leetcode.com/topic/17273/18-22-lines-c-bfs-dfs-solutions

https://discuss.leetcode.com/topic/13854/easy-bfs-topological-sort-java

https://discuss.leetcode.com/topic/25964/ac-python-topological-sort-52-ms-solution-o-v-e-time-and-o-v-e-space

更多討論

https://discuss.leetcode.com/category/215/course-schedule

207. Course Schedule